Explicit homeomorphism between the plane and the plane minus a closed half-line.
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Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$
I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$
I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.
general-topology
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add a comment |
$begingroup$
Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$
I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$
I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.
general-topology
$endgroup$
$begingroup$
${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
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– Paul Frost
Dec 27 '18 at 22:48
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Yes, I have corrected the mistake.
$endgroup$
– MatP
Dec 28 '18 at 4:56
add a comment |
$begingroup$
Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$
I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$
I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.
general-topology
$endgroup$
Let $r$ be a closed half-line in $mathbb{R}^2$ and $delta$ a positive real number. I want to find an explicit homeomorphism $phi:mathbb{R}^2tomathbb{R}^2-r$ such that $phi(X)=X$ for all $X$ such that $d(X,r)geqdelta.$
I can suppose the half-line is ${(x,y)inmathbb{R}^2:xgeq0,y=0}.$ Moreover, if we put $$B={Xinmathbb{R}^2:d(X,r)<delta},$$ we must have $phi(B)=B-r.$
I don’t know how to write this homeomorphism. I thought I can deform this initial configuration into something other, but I don’t have any other ideas.
general-topology
general-topology
edited Dec 28 '18 at 4:56
MatP
asked Dec 27 '18 at 18:00
MatPMatP
1367
1367
$begingroup$
${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
$endgroup$
– Paul Frost
Dec 27 '18 at 22:48
$begingroup$
Yes, I have corrected the mistake.
$endgroup$
– MatP
Dec 28 '18 at 4:56
add a comment |
$begingroup$
${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
$endgroup$
– Paul Frost
Dec 27 '18 at 22:48
$begingroup$
Yes, I have corrected the mistake.
$endgroup$
– MatP
Dec 28 '18 at 4:56
$begingroup$
${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
$endgroup$
– Paul Frost
Dec 27 '18 at 22:48
$begingroup$
${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
$endgroup$
– Paul Frost
Dec 27 '18 at 22:48
$begingroup$
Yes, I have corrected the mistake.
$endgroup$
– MatP
Dec 28 '18 at 4:56
$begingroup$
Yes, I have corrected the mistake.
$endgroup$
– MatP
Dec 28 '18 at 4:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let us proceed in various steps.
(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.
(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.
(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$
It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.
(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$
$endgroup$
$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20
add a comment |
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Using polar coordinates (r,t) where
{ (r,0) : 0 <= r } is the removed closed ray,
map (r,t) to (r,$pi$ + t/2).
$endgroup$
$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
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– Paul Frost
Dec 28 '18 at 9:27
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You got it. Now finish the homeomorphism.
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– William Elliot
Dec 28 '18 at 22:06
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Yes, but it will be not a homeomorphism as required in the question.
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– Paul Frost
Dec 28 '18 at 22:48
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us proceed in various steps.
(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.
(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.
(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$
It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.
(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$
$endgroup$
$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20
add a comment |
$begingroup$
Let us proceed in various steps.
(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.
(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.
(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$
It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.
(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$
$endgroup$
$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20
add a comment |
$begingroup$
Let us proceed in various steps.
(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.
(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.
(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$
It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.
(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$
$endgroup$
Let us proceed in various steps.
(1) It clearly suffices to show that for any $a > 0$ there exists a homeomorphism $h_a : left( [-a,infty) times [-a,a] right) setminus r to [-a,infty) times [-a,a]$ which is the identity on the boundary $B_a$ of the strip $S_a = [-a,infty) times [-a,a]$.
(2) It suffices to consider $a = 1$. Define $g_a : [-1,infty) times [-1,1] to [-a,infty) times [-a,a], g_a(x,y) = (ax,ay)$. This is a homeomorphism (its inverse is $g_{1/a}$) such that $g_a(r) = r$ and $g_a(B_1) = B_a$. If we have found $h_1$, then take $h_a = g_a circ h_1 circ g_{1/a}$.
(3) Define a map $p: [-1,1] times [-1,1] to [-1,1] times [-1,1]$ as follows. For each $y in [-1,1]$ let $p_y : [-1,1] to [-1,1]$ denote the piecewise linear map such that $p_y(-1) = -1$, $p_y(-lvert y rvert) = 1 - 2 lvert y rvert$, $p_y(1) = 1$. Then define $p(x,y) = (p_y(x),y)$. Explicitly
$$p(x,y) =
begin{cases}
1 +2 x & -1 le x le - lvert y rvert \
1 - 2 lvert y rvert + dfrac{2 lvert y rvert}{1 + lvert y rvert}(x + lvert y rvert) & - lvert y rvert le x le 1
end{cases}
$$
It is an easy exercise to verify that $p$ is continuous. Moreover, $p$ is the identity on the boundary $D$ of the square $Q = [-1,1] times [-1,1]$. It maps $[0,1] times { 0 }$ to the point $(0,1)$ and restricts to a continuous bijection $p ': Q' setminus r' to Q'$, where $Q' = [-1,1) times [-1,1]$ and $r' = [0,1) times { 0 }$. Since $Q$ is compact, $p$ is a closed map and therefore a quotient map. Let $U subset Q' setminus r'$ be open. Since $Q' setminus r'$ is open in $Q$, $U$ is open in $Q$. We have $p^{-1}(p(U)) = U$, hence $p(U)$ is open in $Q$. Thus $p'(U) = p(U) = p(U) cap Q'$ is open in $Q'$ which shows that $p'$ is an open map, i.e. a homeomorphism.
(4) Let us finally construct the homeomorphism $h_1$ as in (2). Define a homeomorphism $psi : [-1,1) to [-1,infty)$ by $psi(x) = x$ for $x le 0$ and $psi(x) = dfrac{x}{1-x}$ for $x ge 0$. Then $psi' = psi times id : Q' to S_1$ is a homeomorphism such that $psi'(r') = r$. It maps the boundary of $Q'$ onto the boundary of $S_1$ and restricts to a homeomorphism $phi: Q' setminus r' to S_1 setminus r$. Now define
$$h_1 = phi circ p' circ phi^{-1} .$$
edited Dec 28 '18 at 11:34
answered Dec 28 '18 at 11:28
Paul FrostPaul Frost
11.6k3935
11.6k3935
$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20
add a comment |
$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20
$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20
$begingroup$
@MatP Any reaction would be nice. If you are not satisfied with an answer, you should tell it to author, and you can also downvote. On the other hand, if the answer is helpful, you should accept it. See math.stackexchange.com/help/someone-answers. No reaction also leaves the question in the uanswered queue which is confusing for the users. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,857 questions with no upvoted or accepted answers ". If you look at these questions, you will see that a great many are actually answered.
$endgroup$
– Paul Frost
Jan 7 at 14:20
add a comment |
$begingroup$
Using polar coordinates (r,t) where
{ (r,0) : 0 <= r } is the removed closed ray,
map (r,t) to (r,$pi$ + t/2).
$endgroup$
$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
$endgroup$
– Paul Frost
Dec 28 '18 at 9:27
$begingroup$
You got it. Now finish the homeomorphism.
$endgroup$
– William Elliot
Dec 28 '18 at 22:06
$begingroup$
Yes, but it will be not a homeomorphism as required in the question.
$endgroup$
– Paul Frost
Dec 28 '18 at 22:48
add a comment |
$begingroup$
Using polar coordinates (r,t) where
{ (r,0) : 0 <= r } is the removed closed ray,
map (r,t) to (r,$pi$ + t/2).
$endgroup$
$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
$endgroup$
– Paul Frost
Dec 28 '18 at 9:27
$begingroup$
You got it. Now finish the homeomorphism.
$endgroup$
– William Elliot
Dec 28 '18 at 22:06
$begingroup$
Yes, but it will be not a homeomorphism as required in the question.
$endgroup$
– Paul Frost
Dec 28 '18 at 22:48
add a comment |
$begingroup$
Using polar coordinates (r,t) where
{ (r,0) : 0 <= r } is the removed closed ray,
map (r,t) to (r,$pi$ + t/2).
$endgroup$
Using polar coordinates (r,t) where
{ (r,0) : 0 <= r } is the removed closed ray,
map (r,t) to (r,$pi$ + t/2).
answered Dec 28 '18 at 7:24
William ElliotWilliam Elliot
8,7222820
8,7222820
$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
$endgroup$
– Paul Frost
Dec 28 '18 at 9:27
$begingroup$
You got it. Now finish the homeomorphism.
$endgroup$
– William Elliot
Dec 28 '18 at 22:06
$begingroup$
Yes, but it will be not a homeomorphism as required in the question.
$endgroup$
– Paul Frost
Dec 28 '18 at 22:48
add a comment |
$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
$endgroup$
– Paul Frost
Dec 28 '18 at 9:27
$begingroup$
You got it. Now finish the homeomorphism.
$endgroup$
– William Elliot
Dec 28 '18 at 22:06
$begingroup$
Yes, but it will be not a homeomorphism as required in the question.
$endgroup$
– Paul Frost
Dec 28 '18 at 22:48
$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
$endgroup$
– Paul Frost
Dec 28 '18 at 9:27
$begingroup$
In polar coordinates $(r,t)$ stands for $re^{it}$ with $t in (0,2pi)$. But then $pi +t/2 in (pi,2pi)$, i.e. the image of your map is the open lower half plane.
$endgroup$
– Paul Frost
Dec 28 '18 at 9:27
$begingroup$
You got it. Now finish the homeomorphism.
$endgroup$
– William Elliot
Dec 28 '18 at 22:06
$begingroup$
You got it. Now finish the homeomorphism.
$endgroup$
– William Elliot
Dec 28 '18 at 22:06
$begingroup$
Yes, but it will be not a homeomorphism as required in the question.
$endgroup$
– Paul Frost
Dec 28 '18 at 22:48
$begingroup$
Yes, but it will be not a homeomorphism as required in the question.
$endgroup$
– Paul Frost
Dec 28 '18 at 22:48
add a comment |
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$begingroup$
${(x,y)inmathbb{R}^2:xgeq0}$ is the closed right half plane. Do you mean ${(x,y)inmathbb{R}^2:xgeq0 , y =0}$?
$endgroup$
– Paul Frost
Dec 27 '18 at 22:48
$begingroup$
Yes, I have corrected the mistake.
$endgroup$
– MatP
Dec 28 '18 at 4:56